Branch, Loop and Node Analyses - Engineering

Find I₁. 4I₁ + 4I₂ = 2
6I₁ + 7I₂ = 4

A 0.5 A

B 50 mA

C –0.5 A

D –50 mA

ANS:C - –0.5 A

let us take 4I1+4I2=2.....eqn. no (1) & 6I1+7I2=4....eqn. no (2)
multiply the eqn no 1 by 7 and eqn no 2 by 4 then the eqns will be as follows 28I1+28I2=14....eqn no (3) & 24I1+28I2=16....eqn no (4)
solve eqns 3 & 4 den u get 4I1= -2
therfore I1= -0.5A

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